Let $h(x)=2x^3+3x^2-12x+5$. The absolute minimum value of $h$ over the closed interval $-3 \leq x \leq 2$ occurs at what $x$ -value? Choose 1 answer: Choose 1 answer: (Choice A) A $-2$ (Choice B) B $-3$ (Choice C) C $2$ (Choice D) D $1$
Explanation: $h$ is continuous for all real numbers. Therefore, according to the Extreme Value Theorem, it must have an absolute maximum (or minimum) over any closed interval. Let's first find the relative extremum points of $h$, and then compare them to the function's values on the edges of the interval. The lowest relative minimum will be the function's absolute minimum value. We start with finding the critical points of $h$. The derivative of $h$ is $h'(x)=6(x+2)(x-1)$. $h'(x)=0$ for $x=-2,1$. Since $h'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=-2$ and $x=1$. They are both within the closed interval $-3 \leq x \leq 2$. Our critical points divide the closed interval into three intervals: $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $\llap{-}3 <x< \llap{-}2$ $\llap{-}2<x<1$ $1<x<2$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $-3<x<-2$ $x=-\dfrac52$ $h'\left(-\dfrac52\right)=\dfrac{21}{2}>0$ $h$ is increasing $\nearrow$ $-2<x<1$ $x=0$ $h'\left(0\right)=-12<0$ $h$ is decreasing $\searrow$ $1<x<2$ $x=\dfrac32$ $h'\left(\dfrac32\right)=\dfrac{21}{2}>0$ $h$ is increasing $\nearrow$ Now let's look at all the critical points and the endpoints of the interval: $x$ $h(x)$ Before After Verdict $-3$ $14$ $-$ $\nearrow$ Minimum $-2$ $25$ $\nearrow$ $\searrow$ Maximum $1$ $-2$ $\searrow$ $\nearrow$ Minimum $2$ $9$ $\nearrow$ $-$ Maximum We can see that the absolute minimum point of $h$ is $(1,-2)$, which means the absolute minimum value of $h$ occurs at $x=1$. [I want to see the analysis of all extremum points.] In conclusion, the absolute minimum value of $h$ over $-3\leq x\leq 2$ occurs at $x=1$.